---
title: "Charges, Currents and Conduction"
book: "University Physics — Year 2"
subject: physics
language: en
chapter: 10
exercises: 12
source: https://one-course.com/books/physics/4/en/chapter/10-charges-currents-and-conduction
---

# Chapter 10 — Charges, Currents and Conduction

Flip the switch and the lamp lights at once; yet the electrons in the copper wire drift toward it at a fraction of a millimetre per second, and would take hours to arrive. A lightning bolt carries thirty thousand amperes through a channel the width of a finger; the nerve that moves that finger carries picoamperes. The Year 1 volume treated currents as numbers in wires, $I = U/R$; this chapter describes them as a *field* — a [current density](#def-b2-charges-currents-conduction-densities) at every point of matter — writes the law that charge is never created or destroyed, and derives Ohm’s law from the jostling of electrons in a metal: the microscopic picture behind the resistor, the fuse and the Hall sensor.

![Copper, the conductor of the electrical world: in the strands of this cable some 1029 electrons per cubic metre drift at a fraction of a millimetre per second.](https://one-course.com/images/onecourse/chapters/physics-4/b2-charges-currents-conduction/img-46708c3f1227.jpg)

*Copper, the conductor of the electrical world: in the strands of this cable some $10^{29}$ electrons per cubic metre drift at a fraction of a millimetre per second.*

## 10.1 Charge and current densities

**Definition 10.1 (Charge density; current density).**

At the [mesoscopic scale](https://one-course.com/books/physics/4/en/chapter/2-fluid-kinematics#def-b2-fluid-kinematics-particle), matter carries a *volume charge density* $\rho(M, t)$ ($\mathrm{C}/\mathrm{m}^{3}$): the charge $\rho\,\dd\tau$ in $\dd\tau$. A surface may carry a *surface density* $\sigma$ ($\mathrm{C}/\mathrm{m}^{2}$), a wire a linear density $\lambda$ ($\mathrm{C}/\mathrm{m}$). If the carriers of species $i$ (charge $q_i$, number density $n_i$) move at the mean velocity $\vect v_i$, the *current density* is

$$
\vect j = \sum_in_iq_i\,\vect v_i \qquad (\mathrm{A}/\mathrm{m}^{2}) ,
$$

and the *intensity* through an oriented surface $S$ is the flux $I = \iint_S\vect j\cdot\vect n\,\dd S$: the charge crossing $S$ per unit time. For a single species, $\vect j = \rho_m\vect v$ with $\rho_m = nq$ the density of mobile charge.

**Proof.** In $\dd t$ the carriers of species $i$ that cross $\dd S$ are those in the oblique cylinder of base $\dd S$ and generator $\vect v_i\dd t$: number $n_i\vect v_i\cdot\vect n\,\dd S\,\dd t$, charge $q_i$ times that. ∎

**Example 10.2 (How slowly electrons drift).**

Copper has one conduction electron per atom: $n = \rho_{\text{Cu}}N_A/M =
8900 \times 6.02 \times 10^{23}/0.0635 = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$. A current of $10\,\mathrm{A}$ in a $1.5\,\mathrm{mm}^{2}$ wire is $j = 6.7 \times 10^{6}\,\mathrm{A}/\mathrm{m}^{2}$, and the *[drift velocity](#ex-b2-charges-currents-conduction-drift)* is $v = j/ne = 6.7 \times 10^6/(8.5 \times 10^{28} \times 1.6 \times
10^{-19}) = 0.5\,\mathrm{mm}/\mathrm{s}$ — two hours per metre. The lamp lights at once because the *field* that pushes the electrons is set up along the whole wire within nanoseconds (it travels at nearly the speed of light, as the cable of [Chapter 8](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#ch-b2-dispersion-wave-packets) showed): all the electrons start together, like water in a full hose.

![Left: a tube of current — in a stationary regime the same intensity crosses every section. Right: an electron in a metal zigzags between collisions at about 106 m/s; the field adds a slow drift, opposite to E, of a fraction of a millimetre per second.](https://one-course.com/images/onecourse/chapters/physics-4/b2-charges-currents-conduction/fig-651a78f95299.svg)

*Left: a tube of current — in a stationary regime the same intensity crosses every section. Right: an electron in a metal zigzags between collisions at about $10^6$ m/s; the field adds a slow drift, opposite to $\vect E$, of a fraction of a millimetre per second.*

**Example 10.3 (Orders of magnitude).**

Lightning, $30\,\mathrm{kA}$ in a channel of $1\,\mathrm{cm}$ radius: $j \approx 1 \times 10^{8}\,\mathrm{A}/\mathrm{m}^{2}$. A household wire: $10^6$–$10^7$. An electron beam in a cathode-ray tube, $1\,\mathrm{mA}$ over $1\,\mathrm{mm}^{2}$: $10^3$. A nerve fibre, $1\,\mathrm{nA}$ through $10\,\text{µ}\mathrm{m}^{2}$: $10^2$. The beam of a particle accelerator, $1\,\mathrm{A}$ in a $0.1\,\mathrm{mm}^{2}$ spot: $10^7$ — in vacuum, with no collisions at all.

## 10.2 Conservation of charge

**Theorem 10.4 (Local conservation of charge).**

At every point,

$$
\frac{\partial\rho}{\partial t} + \operatorname{div}\vect j = 0 ;
$$

integrated over a fixed volume $V$ bounded by the closed surface $\Sigma$, $\dd Q_V/\dd t = -I_{\text{out}}$: the charge inside changes only by what crosses the boundary. In a *stationary regime* ($\partial_t\rho = 0$), $\operatorname{div}\vect j = 0$: the flux of $\vect j$ is conserved along a tube of current, the same intensity crosses every section of a wire, and the currents entering a node equal those leaving (Kirchhoff’s node law).

**Proof.** Word for word the mass balance of [Theorem 2.8](https://one-course.com/books/physics/4/en/chapter/2-fluid-kinematics#thm-b2-fluid-kinematics-continuity) with $\rho$ the [charge density](#def-b2-charges-currents-conduction-densities) and $\vect j$ its current: the net outflow of charge through the six faces of a fixed box, per unit volume, is $\operatorname{div}\vect j$, and no charge is created. The integral form sums the boxes (the interior faces cancel). For a tube between two sections $S_1$, $S_2$ with no flux through its side, $I_1 = I_2$; a node is a small volume with several wires: $\sum I_{\text{in}} = \sum I_{\text{out}}$. ∎

**Remark 10.5 (Where the current stops: the capacitor).**

A wire feeding a capacitor plate is a tube of current that ends: the flux of $\vect j$ into the plate is $I = \dd Q/\dd t$, and charge accumulates — $\operatorname{div}\vect j \ne 0$ there, the regime is not stationary. Yet the same current $I$ flows in the wire on the other side: something must carry the balance across the gap. [Chapter 11](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#ch-b2-maxwell-equations) names it — the displacement current — and makes the conservation of charge the keystone of Maxwell’s equations.

## 10.3 The Drude model and Ohm’s law

**Proposition 10.6 (Drude model; local Ohm’s law).**

Model the conduction electrons of a metal as free particles (mass $m$, charge $-e$, density $n$) that feel the field $\vect E$ and, on average, lose their drift momentum by collisions with the lattice every $\tau$ seconds — a [friction force](https://one-course.com/books/physics/4/en/chapter/1-rigid-body-mechanics#def-b2-rigid-body-mechanics-contact) $-m\vect v/\tau$ on the mean velocity:

$$
m\frac{\dd\vect v}{\dd t} = -e\vect E - \frac m\tau\vect v .
$$

In a steady field the drift settles, in a few $\tau$, at $\vect v = -e\tau
\vect E/m$, and the [current density](#def-b2-charges-currents-conduction-densities) is

$$
\vect j = \gamma\,\vect E , \qquad \gamma = \frac{ne^2\tau}{m} :
$$

the *local Ohm’s law*, with $\gamma$ the *conductivity* ($\mathrm{S}/\mathrm{m}$; its inverse $\rho_e = 1/\gamma$ is the *resistivity*, $\Omega\,\mathrm{m}$). The ratio $\mu = e\tau/m$ of the drift speed to the field is the *mobility* ($\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$). For a conductor of length $L$ and uniform section $S$ carrying a uniform $\vect j$, integrating $E = j/\gamma$ along the length gives $U = EL = (L/\gamma S)I$: the resistance $R = L/\gamma S
= \rho_eL/S$ of the Year 1 volume.

**Proof.** With $\vect v = -e\tau\vect E/m$, $\vect j = -ne\vect v = (ne^2\tau/m)\vect E$. The transient $\vect v(t) = \vect v_\infty(1 - \eu^{-t/\tau})$ lasts $\tau \sim
1 \times 10^{-14}\,\mathrm{s}$, instantaneous on every circuit time scale. The resistance follows from $\int\vect E\cdot\dd\vect l$ between the ends. ∎

**Example 10.7 (Copper, from the measured conductivity).**

Copper: $\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$, $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$: $\tau = m\gamma/ne^2 =
9.1 \times 10^{-31} \times 6 \times 10^7/(8.5 \times 10^{28} \times 2.56 \times 10^{-38}) = 2.5 \times 10^{-14}\,\mathrm{s}$ — twenty-five femtoseconds between collisions. The electrons’ random speed in a metal is about $1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ (a quantum effect, not the thermal $\sqrt{3k_BT/m} \approx 1 \times 10^{5}\,\mathrm{m}/\mathrm{s}$), so the mean free path is $\ell
= v\tau \approx 40\,\mathrm{nm}$, a hundred atomic spacings: the electrons do not collide with the atoms themselves (a perfect crystal is transparent to them) but with its defects and thermal vibrations — which is why $\gamma$ falls when a metal is heated or alloyed. Mobility $e\tau/m =
4.4 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$: $4\,\mathrm{mm}/\mathrm{s}$ per volt per metre.

**Proposition 10.8 (Ohm’s law at high frequency).**

For a sinusoidal field $\underline{\vect E}\,\eu^{\iu\omega t}$ the Drude equation gives $\underline{\vect j} = \underline\gamma(\omega)\underline{\vect E}$ with

$$
\underline\gamma(\omega) = \frac{\gamma_0}{1 + \iu\omega\tau} , \qquad \gamma_0 = \frac{ne^2\tau}{m} :
$$

the conductivity is real and equal to its DC value as long as $\omega\tau
\ll 1$, i.e. up to about $1 \times 10^{13}\,\mathrm{Hz}$ for copper — through the whole radio and microwave range; in the infrared and visible ($\omega\tau \gg 1$) it becomes imaginary, $\underline\gamma \approx ne^2/\iu m\omega$: the electrons oscillate freely, out of phase with the field, and the metal behaves as the plasma of [Chapter 14](https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics#ch-b2-waves-in-media).

**Proof.** $\iu\omega m\underline{\vect v} = -e\underline{\vect E} - m\underline{\vect v}/\tau$, whence $\underline{\vect v} = -e\tau\underline{\vect E}/m(1 + \iu\omega\tau)$ and $\underline{\vect j}
= -ne\underline{\vect v}$. ∎

![Left: the Drude conductivity against — real and constant at low frequency, imaginary beyond 1/. Right: the Hall effect in a conducting strip — the carriers are pushed sideways by the field, charge the edges, and the transverse Hall field balances the magnetic force.](https://one-course.com/images/onecourse/chapters/physics-4/b2-charges-currents-conduction/fig-2cfaecd0cb64.svg)

![Left: the Drude conductivity against — real and constant at low frequency, imaginary beyond 1/. Right: the Hall effect in a conducting strip — the carriers are pushed sideways by the field, charge the edges, and the transverse Hall field balances the magnetic force.](https://one-course.com/images/onecourse/chapters/physics-4/b2-charges-currents-conduction/fig-69463e2c3034.svg)

*Left: the Drude conductivity against $\omega\tau$ — real and constant at low frequency, imaginary beyond $1/\tau$. Right: the Hall effect in a conducting strip — the carriers are pushed sideways by the field, charge the edges, and the transverse Hall field balances the magnetic force.*

**Remark 10.9 (Conductors, semiconductors, electrolytes).**

The same law $\vect j = \sum n_iq_i\mu_i\vect E$ holds for every conductor; what changes is $n$ and $\mu$. Metals: $n \sim 10^{29}$, $\gamma \sim 10^7$ S/m. Semiconductors: $n \sim 10^{16}$–$10^{23}$, set by doping and rising steeply with temperature, with two kinds of carriers (electrons and holes) — whence thermistors and diodes. Electrolytes: ions of both signs, $\mu \sim 10^{-8}$–$10^{-7}$ m$^2$/(V s); sea water $\gamma \approx 5\,\mathrm{S}/\mathrm{m}$, tap water $10^{-2}$, pure water $10^{-5}$. Insulators (glass, polymers): $\gamma \lesssim 10^{-12}$ S/m. The Hall effect of the Year 1 volume, in local form $\vect E_H = -\vect j\wedge\vect B/nq$, measures $n$ and the sign of $q$: in aluminium or zinc the carriers turn out to be positive — holes — and no classical model explains that.

## 10.4 Energy: the local Joule law

**Proposition 10.10 (Local Joule law).**

The electric field delivers to the carriers, per unit volume and time, the power $\vect j\cdot\vect E$; in an ohmic conductor this is

$$
p = \vect j\cdot\vect E = \gamma E^2 = \frac{j^2}\gamma \ \ (\mathrm{W}/\mathrm{m}^{3}) ,
$$

all of it turned into heat by the collisions (the drift kinetic energy is negligible and constant). Integrated over a wire, $\int j^2/\gamma\,\dd\tau
= (L/\gamma S)I^2 = RI^2$.

**Proof.** The force $q\vect E$ on each carrier works at the rate $q\vect E\cdot\vect v$; summed over the $n\,\dd\tau$ carriers: $nq\vect v\cdot\vect E\,\dd\tau = \vect j\cdot
\vect E\,\dd\tau$ (the magnetic force does no work). In the Drude steady state this power equals that dissipated by the friction term, i.e. transferred to the lattice. ∎

**Example 10.11 (A wire heating).**

The $1.5\,\mathrm{mm}^{2}$ wire at $10\,\mathrm{A}$: $p = j^2/\gamma = (6.7 \times 10^6)^2/6 \times 10^7 =
7.4 \times 10^{5}\,\mathrm{W}/\mathrm{m}^{3}$, i.e. $1.1\,\mathrm{W}$ per metre — easily evacuated by the air around it. Left to itself (no cooling), copper ($c = 385\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$, $\rho = 8900\,\mathrm{kg}/\mathrm{m}^{3}$) would warm at $p/\rho c = 0.2\,\mathrm{K}/\mathrm{s}$. At $100\,\mathrm{A}$ in the same wire the rate is a hundred times larger, $20\,\mathrm{K}/\mathrm{s}$: it melts in a minute — the principle of the fuse, whose thin wire is designed to melt first.

**Proposition 10.12 (Relaxation of charge in a conductor).**

In an ohmic conductor a volume [charge density](#def-b2-charges-currents-conduction-densities) decays as $\rho(t) =
\rho_0\eu^{-t/\tau_r}$ with

$$
\tau_r = \frac{\varepsilon_0}{\gamma} :
$$

$1.5 \times 10^{-19}\,\mathrm{s}$ for copper, $2 \times 10^{-12}\,\mathrm{s}$ for sea water, $10\,\mathrm{s}$ for glass. Up to frequencies of order $1/\tau_r$ a conductor carries *no volume charge*: any excess charge runs to its surface. The word "conductor" thus depends on the time scale: sea water is a conductor for radio waves and an insulator for visible light.

**Proof.** Charge conservation with $\vect j = \gamma\vect E$ and Gauss’s law in local form, $\operatorname{div}\vect E = \rho/\varepsilon_0$ ([Chapter 11](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#ch-b2-maxwell-equations)): $\partial_t\rho = -\gamma\operatorname{div}\vect E = -(\gamma/\varepsilon_0)\rho$. ∎

**Method 10.13 (From the local to the integral law).**

To find the resistance of a conductor of arbitrary shape: (1) use the symmetry to write $\vect j$ (stationary: its flux is conserved, so in a tube $j \times$ section $= I$); (2) $\vect E = \vect j/\gamma$; (3) integrate $\vect E$ along a line from one electrode to the other: $U = \int\vect E\cdot
\dd\vect l$, and $R = U/I$; (4) check with the Joule power $\int j^2/\gamma\,
\dd\tau = RI^2$. The same steps give the leakage of a cable, the resistance of the ground around a lightning rod, or of a conical contact.

## 10.5 Exercises

**Exercise 10.1 ★.**

A $2.5\,\mathrm{mm}^{2}$ copper wire carries $16\,\mathrm{A}$. [Current density](#def-b2-charges-currents-conduction-densities), drift speed ($n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$), time for an electron to travel the $20\,\mathrm{m}$ to a socket; electric field in the wire ($\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$), voltage drop over $20\,\mathrm{m}$ and power lost.

**Solution of Exercise 10.1.**

$j = 16/2.5 \times 10^{-6} = 6.4 \times 10^{6}\,\mathrm{A}/\mathrm{m}^{2}$; $v = j/ne = 0.47\,\mathrm{mm}/\mathrm{s}$; $20\,\mathrm{m}$ in $4.3 \times 10^4$ s, twelve hours. $E = j/\gamma = 0.11\,\mathrm{V}/\mathrm{m}$; $U = 2.1\,\mathrm{V}$; $P = UI = 34\,\mathrm{W}$.

**Exercise 10.2 ★.**

Current densities: a lightning stroke ($30\,\mathrm{kA}$, radius $1\,\mathrm{cm}$); an electron beam ($2\,\mathrm{mA}$, $0.5\,\mathrm{mm}^{2}$); a nerve axon ($1\,\mathrm{nA}$, radius $2\,\text{µ}\mathrm{m}$); the Earth’s magnetosphere ring current ($1\,\mathrm{MA}$ through about $1 \times 10^{14}\,\mathrm{m}^{2}$). Which exceed the wire of the previous exercise?

**Solution of Exercise 10.2.**

Lightning $3 \times 10^4/\pi10^{-4} = 1 \times 10^{8}\,\mathrm{A}/\mathrm{m}^{2}$; beam $4 \times 10^{3}\,\mathrm{A}/\mathrm{m}^{2}$; axon $10^{-9}/1.3 \times 10^{-11} = 80\,\mathrm{A}/\mathrm{m}^{2}$; ring current $1 \times 10^{-8}\,\mathrm{A}/\mathrm{m}^{2}$. Only the lightning exceeds the wire’s $6 \times 10^6$.

**Exercise 10.3 ★.**

Resistivities at $20{}^{\circ}\mathrm{C}$: copper $1.7 \times 10^{-8}\,\Omega\,\mathrm{m}$, aluminium $2.8 \times 10^{-8}\,\Omega\,\mathrm{m}$, iron $1.0 \times 10^{-7}\,\Omega\,\mathrm{m}$, nichrome $1.1 \times 10^{-6}\,\Omega\,\mathrm{m}$. Resistance of $100\,\mathrm{m}$ of $2.5\,\mathrm{mm}^{2}$ in each; section of aluminium equivalent to $2.5\,\mathrm{mm}^{2}$ of copper, and the mass ratio (densities $8900$ and $2700\,\mathrm{kg}/\mathrm{m}^{3}$); length of $0.5\,\mathrm{mm}$ nichrome wire for a $1\,\mathrm{kW}$, $230\,\mathrm{V}$ heater.

**Solution of Exercise 10.3.**

$R = \rho_eL/S$: $0.68$, $1.1$, $4.0$, $44\,\Omega$. Aluminium: $2.5 \times 2.8/1.7 =
4.1\,\mathrm{mm}^{2}$, mass ratio $(4.1 \times 2700)/(2.5 \times 8900) = 0.50$ — half the mass. Heater: $R = 230^2/1000 = 53\,\Omega$, $S = 0.196\,\mathrm{mm}^{2}$, $L = RS/\rho_e = 9.4\,\mathrm{m}$.

**Exercise 10.4 ★.**

Drude: $\tau$, mobility and mean free path (random speed $1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s}$) for copper ($\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$, $n = 8.5 \times 10^{28}\,\mathrm{m}^{-3}$) and silver ($\gamma = 6.3 \times 10^{7}\,\mathrm{S}/\mathrm{m}$, $n = 5.9 \times 10^{28}\,\mathrm{m}^{-3}$); conductivity of a silicon sample with $n = 1 \times 10^{22}\,\mathrm{m}^{-3}$ and $\mu = 0.14\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$; drift speed in that sample at $1\,\mathrm{kV}/\mathrm{m}$.

**Solution of Exercise 10.4.**

$\tau = m\gamma/ne^2$: copper $2.5 \times 10^{-14}\,\mathrm{s}$, $\mu = e\tau/m = 4.4 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$, $\ell = 40\,\mathrm{nm}$; silver $3.8 \times 10^{-14}\,\mathrm{s}$, $6.7 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$, $61\,\mathrm{nm}$. Silicon: $\gamma = ne\mu = 10^{22} \times 1.6 \times 10^{-19} \times 0.14 = 220\,\mathrm{S}/\mathrm{m}$; $v =
\mu E = 140\,\mathrm{m}/\mathrm{s}$.

**Exercise 10.5 ★★.**

A capacitor of plates $S$ is charged through a wire by $I(t)$. (a) Apply the integral conservation law to a closed surface enclosing one plate: relate $I$ and the plate charge $Q$. (b) Why is $\operatorname{div}\vect j \ne 0$ at the plate? (c) A spherical electrode of radius $a$ in a conducting medium emits a radial current $I$: $\vect j(r)$; check $\operatorname{div}
\vect j = 0$ for $r > a$ (use $\operatorname{div}\vect j = \frac1{r^2}\partial_r(r^2j_r)$ for a radial field). (d) Where, then, does the charge conservation "break"?

**Solution of Exercise 10.5.**

(a) Current in, none out: $\dd Q/\dd t = I$. (b) Charge accumulates: $\partial_t
\rho \ne 0$. (c) $\vect j = (I/4\pi r^2)\vect e_r$: $\operatorname{div}\vect j = \frac1{r^2}
\partial_r(I/4\pi) = 0$. (d) Only at the electrode, where the current is injected from the wire — a source of the flux.

**Exercise 10.6 ★★.**

*Leakage of a [coaxial cable](https://one-course.com/books/physics/4/en/chapter/8-dispersion-and-wave-packets#prop-b2-dispersion-wave-packets-coax).* The insulator between the conductors (radii $a = 1\,\mathrm{mm}$, $b = 3.5\,\mathrm{mm}$) has $\gamma = 1 \times 10^{-13}\,\mathrm{S}/\mathrm{m}$; a voltage $U$ is applied between them, a radial leakage current $I$ flows per length $L$. (a) $\vect j(r)$ and $\vect E(r)$. (b) Show that the resistance of a length $L$ is $R = \ln(b/a)/2\pi\gamma L$; value per kilometre. (c) Leakage current and power per kilometre at $1\,\mathrm{kV}$. (d) Compare with the capacitance per length $2\pi\varepsilon_0\varepsilon_r/\ln(b/a)$: show $RC = \varepsilon_0\varepsilon_r/\gamma$ and interpret with the relaxation time.

**Solution of Exercise 10.6.**

(a) $j = I/2\pi rL$, $E = j/\gamma$, radial. (b) $U = \int_a^bE\,\dd r = I\ln(b/a)/
2\pi\gamma L$: $R = \ln3.5/(2\pi \times 10^{-13} \times 10^3) = 2 \times 10^{9}\,\Omega$ per kilometre. (c) $0.5\,\text{µ}\mathrm{A}$, $0.5\,\mathrm{mW}$. (d) $RC = \varepsilon_0\varepsilon_r/\gamma = 200\,\mathrm{s}$ for $\varepsilon_r = 2.3$: the cable, charged and isolated, discharges through its own insulator with the relaxation time of that insulator.

**Exercise 10.7 ★★.**

[Complex conductivity](#prop-b2-charges-currents-conduction-acdrude) of copper ($\tau = 2.5 \times 10^{-14}\,\mathrm{s}$). (a) $\omega\tau$ at $50\,\mathrm{Hz}$, $10\,\mathrm{GHz}$ (microwaves), $30\,\mathrm{THz}$ (infrared), $5 \times 10^{14}\,\mathrm{Hz}$ (green light). (b) Modulus and phase of $\underline\gamma/\gamma_0$ in each case. (c) Up to what frequency is Ohm’s law good to $1\%$? (d) In the visible, write $\underline\gamma$ and show that $\underline{\vect j}$ is in quadrature with $\underline{\vect E}$: mean Joule power?

**Solution of Exercise 10.7.**

(a) $\omega\tau$: $8 \times 10^{-12}$, $1.6 \times 10^{-3}$, $4.7$, $79$. (b) $|\underline\gamma|/\gamma_0 =
1/\sqrt{1 + \omega^2\tau^2}$: $1$, $1$, $0.21$, $0.013$; phase $-\arctan\omega\tau$: $0$, $-0.09{}^{\circ}$, $-78{}^{\circ}$, $-89{}^{\circ}$. (c) $\omega\tau \le 0.14$: $f \le
0.9\,\mathrm{THz}$. (d) $\underline\gamma \approx ne^2/\iu m\omega$: $\underline{\vect j}$ lags $\underline{\vect E}$ by $90{}^{\circ}$, $\langle\vect j\cdot\vect E\rangle = 0$: no Joule heating — the electrons oscillate freely, as in a plasma.

**Exercise 10.8 ★★.**

A $0.20\,\mathrm{mm}$ copper fuse wire (copper: $\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$, $\rho =
8900\,\mathrm{kg}/\mathrm{m}^{3}$, $c = 385\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$, melts at $1085{}^{\circ}\mathrm{C}$, latent heat $205\,\mathrm{kJ}/\mathrm{kg}$). (a) [Current density](#def-b2-charges-currents-conduction-densities) and Joule power per unit volume at $10\,\mathrm{A}$. (b) Time to melt from $20{}^{\circ}\mathrm{C}$ if no heat escapes (ignore the rise of resistivity with temperature). (c) Same at $30\,\mathrm{A}$. (d) In reality the wire loses heat to the air: how does that set the rated current of a fuse, and why are fuses sealed in a sand-filled cartridge?

**Solution of Exercise 10.8.**

(a) $S = 3.1 \times 10^{-8}\,\mathrm{m}^{2}$, $j = 3.2 \times 10^{8}\,\mathrm{A}/\mathrm{m}^{2}$, $p = j^2/\gamma = 1.7 \times 10^{9}\,\mathrm{W}/\mathrm{m}^{3}$. (b) Energy to melt per unit volume $\rho(c\Delta T + L_f) = 8900(4.1 \times 10^5 + 2.05 \times
10^5) = 5.5 \times 10^{9}\,\mathrm{J}/\mathrm{m}^{3}$: $3.2\,\mathrm{s}$. (c) Nine times faster: $0.36\,\mathrm{s}$. (d) The rated current is the one whose Joule heating the air just evacuates below the melting point; the sand quenches the arc that would otherwise keep conducting across the gap after the wire melts.

**Exercise 10.9 ★★.**

*Relaxation.* (a) Derive $\partial_t\rho = -(\gamma/\varepsilon_0)\rho$ from charge conservation, Ohm’s law and $\operatorname{div}\vect E = \rho/\varepsilon_0$. (b) $\tau_r$ for copper, sea water ($\gamma = 5\,\mathrm{S}/\mathrm{m}$), tap water ($5 \times 10^{-2}\,\mathrm{S}/\mathrm{m}$), glass ($1 \times 10^{-12}\,\mathrm{S}/\mathrm{m}$), PTFE ($1 \times 10^{-16}\,\mathrm{S}/\mathrm{m}$). (c) Up to what frequency is each a "good conductor" (charge relaxing within a period)? (d) A charged glass rod keeps its charge for minutes: is that consistent?

**Solution of Exercise 10.9.**

(a) $\partial_t\rho = -\operatorname{div}\vect j = -\gamma\operatorname{div}\vect E = -\gamma\rho/
\varepsilon_0$. (b) $\varepsilon_0/\gamma$: $1.5 \times 10^{-19}\,\mathrm{s}$, $1.8 \times 10^{-12}\,\mathrm{s}$, $1.8 \times 10^{-10}\,\mathrm{s}$, $9\,\mathrm{s}$, $9 \times 10^4$ s. (c) $f \lesssim 1/2\pi\tau_r$: $10^{18}$ Hz (the [Drude model](#prop-b2-charges-currents-conduction-drude) fails long before), $90\,\mathrm{GHz}$, $0.9\,\mathrm{GHz}$, $0.02\,\mathrm{Hz}$, $2 \times 10^{-6}\,\mathrm{Hz}$. (d) A glass rod would lose its charge in a minute through its volume; real dry glass is nearer $10^{-14}$ S/m (hours), and it is surface humidity that usually discharges it — consistent in order of magnitude.

**Exercise 10.10 ★★★.**

*Grounding.* A hemispherical electrode of radius $a = 0.50\,\mathrm{m}$ is buried flush with the surface of soil of conductivity $\gamma =
0.010\,\mathrm{S}/\mathrm{m}$; a current $I$ enters it and spreads radially into the half-space. (a) $\vect j(r)$, $\vect E(r)$, potential $V(r)$ with $V(\infty) = 0$. (b) Resistance of the grounding, $R = 1/2\pi\gamma a$; value. (c) A lightning stroke of $20\,\mathrm{kA}$ hits the rod: potential of the rod; potential at $r = 10\,\mathrm{m}$ and $10.7\,\mathrm{m}$; the "step voltage" between the feet of a person standing there (feet $0.7\,\mathrm{m}$ apart, radially). (d) Why are cattle, with their four legs far apart, killed by nearby strikes more often than people?

**Solution of Exercise 10.10.**

(a) $\vect j = (I/2\pi r^2)\vect e_r$, $E = I/2\pi\gamma r^2$, $V = I/2\pi\gamma r$. (b) $R = V(a)
/I = 1/2\pi\gamma a = 32\,\Omega$. (c) $V_{\text{rod}} = 640\,\mathrm{kV}$; $V(10) = 32\,\mathrm{kV}$, $V(10.7) = 30\,\mathrm{kV}$: $2.1\,\mathrm{kV}$ between the feet. (d) Front and hind legs $1.5\,\mathrm{m}$ apart along the radius: $4$–$5\,\mathrm{kV}$, and the current path crosses the heart.

**Exercise 10.11 ★★★.**

*Hall sensor.* A strip of thickness $t$ and width $w$ carries $I$ along $x$ in a field $B$ along $z$. (a) From $\vect E_H = -\vect j\wedge\vect B
/nq$, show that the Hall voltage across the width is $V_H = IB/nqt$. (b) Copper strip, $t = 0.10\,\mathrm{mm}$, $1\,\mathrm{A}$, $1\,\mathrm{T}$: $V_H$. (c) A doped semiconductor with $n = 1 \times 10^{22}\,\mathrm{m}^{-3}$, same geometry: $V_H$; why sensors use semiconductors. (d) A sensor with $n = 1 \times 10^{22}\,\mathrm{m}^{-3}$, $t =
20\,\text{µ}\mathrm{m}$, fed with $1\,\mathrm{mA}$, reads $10\,\text{µ}\mathrm{V}$: field measured. What does the sign of $V_H$ tell?

**Solution of Exercise 10.11.**

(a) $E_H = jB/nq$ with $j = I/wt$: $V_H = E_Hw = IB/nqt$. (b) $1/(8.5 \times 10^{28}
\times 1.6 \times 10^{-19} \times 10^{-4}) = 0.7\,\text{µ}\mathrm{V}$. (c) $6.3\,\mathrm{mV}$: $V_H \propto 1/n$, and semiconductors have $10^6$ times fewer carriers. (d) $B = V_Hnqt/I =
10^{-5} \times 10^{22} \times 1.6 \times 10^{-19} \times 2 \times 10^{-5}/10^{-3} = 0.32\,\mathrm{T}$; the sign of $V_H$ gives the sign of the carriers (for a known $\vect B$), or the direction of $\vect B$ (for a known sensor).

**Exercise 10.12 ★★★.**

*Electrolyte.* In water, Na$^+$ and Cl$^-$ ions have mobilities $\mu_+ = 5.2 \times 10^{-8}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$ and $\mu_- = 7.9 \times 10^{-8}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$. (a) Show that $\gamma = ne(\mu_+ + \mu_-)$ for $n$ ion pairs per unit volume. (b) Sea water, $0.5\,\mathrm{mol}/\mathrm{L}$ of NaCl: $\gamma$; compare with the measured $5\,\mathrm{S}/\mathrm{m}$. (c) Drift speeds at $10\,\mathrm{V}/\mathrm{m}$; which ion carries more of the current? (d) A column of sea water, $1\,\mathrm{cm}^{2}$ by $10\,\mathrm{cm}$, carries $1\,\mathrm{A}$: voltage, power, and the time to warm it by $10\,\mathrm{K}$ ($c =
4.0\,\mathrm{kJ}/(\mathrm{kg}\,\mathrm{K})$); why electrophoresis gels are cooled.

**Solution of Exercise 10.12.**

(a) $\vect j = ne\mu_+\vect E + (-ne)(-\mu_-\vect E)$: both species drift in opposite directions and carry current in the same direction. (b) $n =
0.5 \times 10^3 \times 6.02 \times 10^{23} = 3 \times 10^{26}\,\mathrm{m}^{-3}$: $\gamma = 3 \times 10^{26} \times 1.6 \times 10^{-19}
\times 1.31 \times 10^{-7} = 6.3\,\mathrm{S}/\mathrm{m}$ (the ions hinder each other at this concentration: $5\,\mathrm{S}/\mathrm{m}$ measured). (c) $0.52\,\text{µ}\mathrm{m}/\mathrm{s}$ and $0.79\,\text{µ}\mathrm{m}/\mathrm{s}$: Cl$^-$ carries $60\%$. (d) $R = L/\gamma S = 200\,\Omega$, $U = 200\,\mathrm{V}$, $P =
200\,\mathrm{W}$ into $10\,\mathrm{g}$ of water: $10\,\mathrm{K}$ in $2\,\mathrm{s}$ — hence the cooling plates of electrophoresis tanks.

## 10.6 Problem: The copper wire, from the atom to the grid

**Problem 10.1.**

Weekend problem — one metre of household wire examined at four scales: the electron, the local laws, the circuit, and the power line

Copper: molar mass $63.5\,\mathrm{g}/\mathrm{mol}$, density $8900\,\mathrm{kg}/\mathrm{m}^{3}$, one conduction electron per atom, conductivity $\gamma = 6.0 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ at $20{}^{\circ}\mathrm{C}$, rising resistivity $\rho_e(T) = \rho_e(20)[1 +
\alpha(T - 20)]$ with $\alpha = 4.0 \times 10^{-3}\,\mathrm{K}^{-1}$; $c = 385\,\mathrm{J}/(\mathrm{kg}\,\mathrm{K})$. Wire: section $S = 2.5\,\mathrm{mm}^{2}$, length $1\,\mathrm{m}$, current $16\,\mathrm{A}$.

**Part I — The electrons.**

1. Number density $n$ of conduction electrons.
2. [Current density](#def-b2-charges-currents-conduction-densities) and [drift velocity](#ex-b2-charges-currents-conduction-drift) ; how long an electron needs to cross the metre.
3. Drude’s $\tau$ and the mobility; mean free path with a random speed of $1.6 \times 10^{6}\,\mathrm{m}/\mathrm{s}$ ; how many atomic spacings ( $0.26\,\mathrm{nm}$ ) is that, and what does it say about what the electrons collide with?
4. The thermal vibrations grow with $T$ , so $\tau \propto 1/T$ roughly: check against $\alpha$ near room temperature ( $\dd\rho_e/\rho_e =  \dd T/T$ would give $\alpha = 1/293$ ).
5. Electric field in the wire; total voltage over the metre; the kinetic energy of the drift per electron compared with $k_BT$ — is the gas of electrons disturbed by the current?
6. Number of electrons crossing a section of the wire per second.
7. Charge of the conduction electrons in the metre of wire; the force the field exerts on all of them together; compare with the weight of the wire. Where does that force end up (think of the collisions)?

**Part II — The local laws.**

8. Joule power per unit volume and per metre.
9. The wire, in air, loses heat at a rate $h(T - T_a)$ per unit surface with $h = 10\,\mathrm{W}/(\mathrm{m}^{2}\,\mathrm{K})$ (convection): equilibrium temperature rise; check whether the resistivity correction matters.
10. The same wire buried in insulation that lets only $h =  2\,\mathrm{W}/(\mathrm{m}^{2}\,\mathrm{K})$ through: new temperature rise, with the resistivity correction included (solve the balance).
11. Left without any cooling, rate of temperature rise at $16\,\mathrm{A}$ and at $160\,\mathrm{A}$ (a short circuit); time to reach the melting point ( $1085{}^{\circ}\mathrm{C}$ ) in the second case; what protects the installation?
12. At $50\,\mathrm{Hz}$ , $\omega\tau = ?$ Is Ohm’s law affected? (The skin effect of [Chapter 14](https://one-course.com/books/physics/4/en/chapter/14-electromagnetic-waves-in-plasmas-conductors-and-dielectrics#ch-b2-waves-in-media) is a different matter.)
13. Write the relaxation time $\varepsilon_0/\gamma$ of copper; what does it imply for the [charge density](#def-b2-charges-currents-conduction-densities) inside the wire, and where does the charge that produces the field $E$ inside the wire sit?

**Part III — Conservation of charge in the circuit.** The wire feeds a capacitor $C = 100\,\text{µ}\mathrm{F}$ through a switch; the current is $I(t) = I_0\eu^{-t/\tau_c}$ with $I_0 = 16\,\mathrm{A}$, $\tau_c = 1\,\mathrm{ms}$.

14. Charge that reaches the plate; final voltage.
15. Apply $\dd Q/\dd t = -I_{\text{out}}$ to a closed surface around the plate: which side carries a conduction current, which does not? What quantity must therefore vary in the gap?
16. At $t = 0.5\,\mathrm{ms}$ , rate of change of the plate charge; if the plates are $1\,\mathrm{cm}^{2}$ apart by $0.1\,\mathrm{mm}$ in air, rate of change of the field between them.
17. Show that $\varepsilon_0\,\dd E/\dd t \times$ plate area equals the current in the wire — the quantity that [Chapter 11](https://one-course.com/books/physics/4/en/chapter/11-maxwells-equations#ch-b2-maxwell-equations) adds to Ampère’s law.
18. The wire is cut and the two ends are $1\,\mathrm{mm}$ apart in air: what current flows (air $\gamma \approx 1 \times 10^{-14}\,\mathrm{S}/\mathrm{m}$ )? At what field does air break down ( $3\,\mathrm{MV}/\mathrm{m}$ ): what voltage across the gap makes a spark?

**Part IV — The line.** A $230\,\mathrm{V}$ line of $10\,\mathrm{mm}^{2}$ copper, $500\,\mathrm{m}$ long (go and return: $1\,\mathrm{km}$ of wire), feeds a farm drawing $40\,\mathrm{A}$.

19. Resistance, voltage drop, and power lost in the line; fraction of the delivered power.
20. The same power at $20\,\mathrm{kV}$ with a transformer at each end: current, drop and loss; conclude why transmission is done at high voltage.
21. Aluminium ( $\gamma = 3.6 \times 10^{7}\,\mathrm{S}/\mathrm{m}$ , $2700\,\mathrm{kg}/\mathrm{m}^{3}$ ) is used on pylons: section for the same resistance as $10\,\mathrm{mm}^{2}$ of copper; mass per kilometre for both; why aluminium wins up there and copper in the wall.
22. A Hall sensor clamped around the line reads its field: for a clamp meter with $n = 1 \times 10^{22}\,\mathrm{m}^{-3}$ , $t = 0.1\,\mathrm{mm}$ , fed with $5\,\mathrm{mA}$ in a $0.1\,\mathrm{T}$ field concentrated by an iron core, the Hall voltage.
23. After a hot day’s load the wire sits at $80{}^{\circ}\mathrm{C}$ : its resistance then, and the extra loss.
24. Lightning hits the line: $20\,\mathrm{kA}$ for $100\,\text{µ}\mathrm{s}$ . Energy dissipated in the $1\,\mathrm{km}$ of wire; temperature rise if none escapes; [current density](#def-b2-charges-currents-conduction-densities) — compare with Part I.
25. Sum up: the five quantities ( $n$ , $\tau$ , $\gamma$ , $j$ , $E$ ) that describe the wire at each scale, and the law that links each pair.

**Solution of Problem 10.1.**

**1.** $n = \rho N_A/M = 8.4 \times 10^{28}\,\mathrm{m}^{-3}$.

**2.** $j = 6.4 \times 10^{6}\,\mathrm{A}/\mathrm{m}^{2}$, $v = j/ne = 0.47\,\mathrm{mm}/\mathrm{s}$; $35\,\mathrm{min}$ per metre.

**3.** $\tau = m\gamma/ne^2 = 2.5 \times 10^{-14}\,\mathrm{s}$, $\mu = 4.4 \times 10^{-3}\,\mathrm{m}^{2}/(\mathrm{V}\,\mathrm{s})$, $\ell =
40\,\mathrm{nm}$, $150$ spacings: the electrons collide with defects and thermal vibrations, not with the atoms.

**4.** $\rho_e \propto T$ gives $\alpha = 1/293 = 3.4 \times 10^{-3}\,\mathrm{K}^{-1}$, close to the measured $4 \times 10^{-3}$.

**5.** $E = 0.11\,\mathrm{V}/\mathrm{m}$, $U = 0.11\,\mathrm{V}$; $\tfrac12mv^2 = 1 \times 10^{-37}\,\mathrm{J}$ against $k_BT = 4 \times 10^{-21}\,\mathrm{J}$: the electron gas does not notice the current.

**6.** $I/e = 1 \times 10^{20}\,$ electrons per second.

**7.** $Q = neSL = 3.4 \times 10^{4}\,\mathrm{C}$; $QE = 3.6\,\mathrm{kN}$, against a weight of $0.2\,\mathrm{N}$. The field pulls the positive lattice with the opposite force; the electrons hand their momentum to the lattice by collisions: the wire as a whole feels nothing.

**8.** $p = j^2/\gamma = 6.8 \times 10^{5}\,\mathrm{W}/\mathrm{m}^{3}$, $1.7\,\mathrm{W}/\mathrm{m}$.

**9.** Surface $\pi d = 5.6 \times 10^{-3}\,\mathrm{m}^{2}$ per metre: $\Delta T = 1.7/(10 \times 5.6 \times
10^{-3}) = 30\,\mathrm{K}$; the resistivity rises $12\%$, the power to $1.9\,\mathrm{W}$: $\Delta T \approx 35\,\mathrm{K}$.

**10.** $\Delta T = 1.7(1 + 0.004\Delta T)/(2 \times 5.6 \times 10^{-3})$: $\Delta T(1 - 0.6) = 150$, $\Delta T \approx 400\,\mathrm{K}$ — near thermal runaway: buried cables must be derated.

**11.** $p/\rho c = 0.2\,\mathrm{K}/\mathrm{s}$; at $160\,\mathrm{A}$, $20\,\mathrm{K}/\mathrm{s}$: melting in about a minute (less as $\rho_e$ rises); the breaker opens in milliseconds.

**12.** $\omega\tau = 8 \times 10^{-12}$: Ohm’s law is untouched.

**13.** $\varepsilon_0/\gamma = 1.5 \times 10^{-19}\,\mathrm{s}$: no charge inside; the field in the wire is made by charges on its *surface*, whose density varies along the wire.

**14.** $Q = I_0\tau_c = 16\,\mathrm{mC}$, $V = Q/C = 160\,\mathrm{V}$.

**15.** The wire side carries $I$, the gap side no conduction current: the charge grows, and with it the electric field in the gap.

**16.** $I = 16\eu^{-0.5} = 9.7\,\mathrm{A}$; $E = Q/\varepsilon_0A$, $\dd E/\dd t = I/\varepsilon_0A
= 9.7/(8.85 \times 10^{-12} \times 10^{-4}) = 1.1 \times 10^{16}\,\mathrm{V}\,\mathrm{m}^{-1}\,\mathrm{s}^{-1}$.

**17.** $\varepsilon_0A\,\dd E/\dd t = I$: the displacement current.

**18.** $j = \gamma E = 10^{-14} \times 2.3 \times 10^5 = 2 \times 10^{-9}\,\mathrm{A}/\mathrm{m}^{2}$, $10^{-14}$ A through the section: nothing; breakdown at $3\,\mathrm{kV}$ across the millimetre.

**19.** $R = 1000/(6 \times 10^7 \times 10^{-5}) = 1.7\,\Omega$; $\Delta U = 67\,\mathrm{V}$; $P =
2.7\,\mathrm{kW}$, $29\%$ of the $9.2\,\mathrm{kW}$ delivered.

**20.** $I = 0.46\,\mathrm{A}$, $\Delta U = 0.8\,\mathrm{V}$, $0.35\,\mathrm{W}$: transmission at high voltage divides the loss by $(U_2/U_1)^2$.

**21.** $10 \times 6/3.6 = 17\,\mathrm{mm}^{2}$; $89\,\mathrm{kg}/\mathrm{km}$ of copper against $45\,\mathrm{kg}/\mathrm{km}$ of aluminium: lighter and cheaper on pylons; in the wall copper’s smaller section fits the conduits and terminals.

**22.** $V_H = IB/nqt = 5 \times 10^{-3} \times 0.1/(10^{22} \times 1.6 \times 10^{-19} \times 10^{-4}) =
3.1\,\mathrm{mV}$.

**23.** $+24\%$: $2.1\,\Omega$, $3.3\,\mathrm{kW}$ lost.

**24.** $RI^2t = 1.7 \times 4 \times 10^8 \times 10^{-4} = 67\,\mathrm{kJ}$ into $89\,\mathrm{kg}$: $2\,\mathrm{K}$. $j = 2 \times 10^{9}\,\mathrm{A}/\mathrm{m}^{2}$, three hundred times the household value.

**25.** $n$ (matter) and $\tau$ (collisions) give $\gamma = ne^2\tau/m$; $\gamma$ links $j$ and $E$ (Ohm); $j$ integrates to $I$, $E$ to $U$ (the circuit); $j^2/\gamma$ is the heat.